Homotopy groups of spheres and low-dimensional topology

Andrew Putman

Abstract

We give a modern account of Pontryagin’s approach to calculating πn+1(Sn) andπn+2(Sn) using techniques from low-dimensional topology.

Contents1 Introduction 12 The degree of a map and πn(Sn) 23 The Pontryagin–Thom construction 44 The Freudenthal suspension theorem 65 Framed circles and πn+1(Sn) 86 Linking numbers 127 The Hopf invariant 138 Framed surgery 159 Improving the Freudenthal suspension theorem 1610 Framed surfaces and πn+2(Sn) 18

1 IntroductionWhen introducing the homotopy groups of spheres, one standard approach deduces manyinitial calculations using techniques from smooth manifolds and geometric topology.

• The fact that the universal cover of S1 is R implies immediately that

πk+1(S1) =⎧⎪⎪⎨⎪⎪⎩

Z if k = 0,

0 if k > 0.

• If 1 ≤ m < n, then an arbitrary continuous map f ∶ Sm → Sn can be homotopedto be smooth and transverse to a point p ∈ Sn. The fact that m < n implies thatf−1(p) = ∅. We deduce that the image of f lies in the contractible set Sn ∖ p, so fis nullhomotopic. The upshot is that πm(Sn) = 0 for 1 ≤m < n.

• Homotopy classes of maps f ∶ Sn → Sn are classified by their degree. Choosing asmooth representative of f , this is the signed count of f−1(p) for a regular valuep ∈ Sn. This leads to the calculation πn(Sn) ≅ Z (see §2 below).

Beyond this, the flavor of the subject changes dramatically. In his thesis, Serre introducedpowerful algebraic tools that have come to dominate homotopy theory. They quickly lead

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to remarkable results and calculations, but the above primitive geometry recedes into thedistance.

However, low-dimensional topology has more to say about homotopy theory than theabove results. The beautiful work of Pontryagin in the 1950’s contained geometric deriva-tions of

πn+1(Sn) =⎧⎪⎪⎨⎪⎪⎩

Z if n = 2,

Z/2 if n ≥ 3

andπn+2(Sn) = Z/2 (n ≥ 2).

This was continued by Rochlin, who managed to prove that πn+3(Sn) ≅ Z/24 for n ≫ 0.The details of these proofs are not well-known today largely because it seems very difficultto push them any further than this.

In these notes, we give a detailed account of the aforementioned calculations of Pon-tryagin. The main source for them is his book [14]. This book is not easy to read todayfor two reasons. First, it is intended also as a textbook on smooth manifold theory, so itspends a huge amount of time developing well-known foundational results from a somewhatarchaic point of view. Second, it avoids using any algebraic topology at all, preferring todevelop things from scratch. As an extreme example of this, in [14, §15B] a complicatedequivalence relation on immersed submanifolds of a surface is given and the equivalenceclasses are called the connectivity group of a surface, which Pontryagin denotes ∆1. In fact,∆1 is actually the first homology group of the surface with Z/2-coefficients, but it is hardto prove this from Pontryagin’s definitions (which do not generalize to spaces other thansurfaces). We do not discuss Rochlin’s later work; see [4] for a modern discussion of it.

The outline of these notes is as follows. As a warm-up, in §2 we discuss the degreeof a map to an n-sphere. In §3 we introduce the Pontryagin–Thom construction. Thesefirst two sections cover material that is discussed in many textbooks, so we only sketchsome of the proofs. In the later sections, our proofs are more complete. In §4, we usethe Pontryagin–Thom construction to prove the Freudenthal suspension theorem. In §5,we calculate πn+1(Sn). In §6, we introduce linking numbers. In §7, we discuss the Hopfinvariant of a map S2n−1 → Sn. In §8, we introduce framed surgery (but to simplify thingswe only discuss how to attach 1-handles). In §9, we show how to use the Hopf invariantto sharpen the Freudenthal suspension theorem. Finally, in §10 we calculate πn+2(Sn).Throughout, we follow the broad outlines of Pontryagin’s book, but our treatment of thedetails is often quite different.

2 The degree of a map and πn(Sn)As a warm-up, this section will discuss the degree of a self-map of Sn, which was introducedby Brouwer [1]. We only sketch the proofs; see [12, §5] for a complete account.

Consider x ∈ πn(Sn). Fix some p ∈ Sn, and realize x by a smooth map f ∶ Sn → Sn thatis transverse to p (such maps clearly exist). Write f−1(p) = q1, . . . , qk. For 1 ≤ i ≤ k, themap on tangent spaces f∗ ∶ (TSn)qi → (TSn)p is an isomorphism; define ϵi = 1 if it preservesthe usual orientation and ϵi = −1 if it reverses it. We will call ϵi the sign of qi. The degree

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of x is

degree(x) ∶=k

∑i=1

ϵi ∈ Z.

Lemma 2.1. For x ∈ πn(Sn), the number degree(x) does not depend on the choice of f orp.

Proof. We first show that it does not depend on f . Let f ′ ∶ Sn → Sn be another smoothrealization of x that is transverse to p. Let F ∶ Sn × [0, 1] → Sn be a homotopy from f tof ′. Homotoping F slightly (without changing F (⋅, 0) or F (⋅, 1)), we can assume that F istransverse to p. The inverse image F−1(p) is then an oriented 1-submanifold of Sn × [0, 1].Each component of F−1(p) is either a circle or an arc connecting two points on Sn × 0, 1.Letting α be such an arc, there are three possibilities.

• The endpoints of α are on Sn × 0. Then α connects two points of f−1(p) whose signsare opposite.

• The endpoints of α are on Sn × 1. Then α connects two points of (f ′)−1(p) whosesigns are opposite.

• One endpoint of α is on Sn × 0 and the other is on Sn × 1. Then α connects a pointof f−1(p) to a point of (f ′)−1(p), and the signs of these points are the same.

The above pairwise matching up of points of f−1(p)∪ (f ′)−1(p) implies that the sum of thesigns of f−1(p) equals the sum of the signs of (f ′)−1(p), as desired.

We now show that degree(x) does not depend on the choice of p. Consider some otherpoint q ∈ Sn, and let τ ∈ SOn+1(R) take p to q. The group SOn+1(R) is connected, so themap τ f ∶ Sn → Sn is homotopic to f (and in particular realizes x). Also, τ f is transverseto q and satisfies (τ f)−1(q) = f−1(p). Finally, the signs of the points of (τ f)−1(q) andthe points of f−1(p) are the same. The desired independence follows.

The mapdegree ∶ πn(Sn)→ Z

is easily seen to be a homomorphism. We now prove that it is an isomorphism. This wasoriginally proved by Hopf [7] following earlier work of Brouwer.

Theorem 2.2. The map degree ∶ πn(Sn)→ Z is an isomorphism for n ≥ 1.

Proof. This is trivial for n = 1, so we will assume that n ≥ 2. It is clear that degree(id) = 1,so degree is surjective. To see that it is injective, consider a smooth map f ∶ Sn → Sn thatis transverse to a point p ∈ Sn and has degree 0. Let f−1(p) = q1, . . . , qk. Since the degreeof f is 0, there exist 1 ≤ i < j ≤ k such that the signs of qi and qj are opposite. It is easy tosee that we can move qi and qj until they collide and “cancel” (see [12] for details; this iswhere we use the fact that n ≥ 2). Doing this repeatedly, we can assume that f−1(p) = ∅,so the image of f lies in the contractible set Sn ∖ p. This implies that f is nullhomotopic,as desired.

Remark. Everything that we discussed above also holds word-for-word for homotopyclasses of maps from a compact orientable Mn to Sn.

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3 The Pontryagin–Thom constructionIn this section, we will describe our main tool for understanding the homotopy groups ofspheres. Fix some n ≥ 1 and k ≥ 0, and let Mk be a k-dimensional submanifold of Rn+k.The restriction to Mk of the tangent bundle of Rn+k is a trivial vector bundle Mk ×Rn+k.The tangent bundle TMk is a subbundle of Mk × Rn+k. Also, the usual Euclidean innerproduct on the tangent bundle of Rn+k induces a bilinear form on each fiber of Mk ×Rn+k.The normal bundle of Mk, denoted NRn+k/Mk , is the fiberwise orthogonal complement inMk ×Rn+k of TMk . The bundle NRn+k/Mk is an n-dimensional vector bundle. A framing onMk is a vector space isomorphism

f ∶Mk ×Rn Ð→ NRn+k/Mk .

Of course, this only exists if NRn+k/Mk is a trivial bundle.A framed k-manifold in Rn+k is a closed k-dimensional submanifold Mk of Rn+k together

with a fixed framing on Mk. A framed cobordism between framed k-manifolds Mk and Nk

is a (k + 1)-dimensional submanifold Ck+1 of Rn+k × [0, 1] ⊂ Rn+k+1 such that

∂Ck+1 = Ck+1 ∩ (Rn+k × 0, 1) = (Mk × 0) ∪ (Nk × 1)

together with a framing on Ck+1 that restricts to the given framings on Mk × 0 and Nk × 1.If such a framed cobordism exists, we will say that Mk and Nk are framed cobordant. Thisdefines an equivalence relation on the set of framed k-manifolds in Rn+k; let Ωfr

k (Rn+k)

denote the set of equivalence classes. The set Ωfrk (R

n+k) is an abelian group under theoperation of disjoint union.

The main result concerning Ωfrk (R

n+k) is the following theorem of Pontryagin [14].

Theorem 3.1. For all n ≥ 1 and k ≥ 0, we have Ωfrk (R

n+k) ≅ πn+k(Sn).

Proof. Since the proof is cogently described in many textbooks (for instance, Milnor’s book[12] and Hirsch’s book [6]), we will only sketch it. The proof will be very similar to thatof Lemma 2.1 discussed above, which shows that the degree is well-defined. It will haveseveral steps.

Step 1. We construct a homomorphism Φ ∶ πn+k(Sn)→ Ωfrk (R

n+k).

Write Sn+k = Rn+k ∪ ∞ and Sn = Rn ∪ ∞. Fix a point p ∈ Rn and an orientation-preserving isomorphism (TRn)p ≅ Rn on the level of tangent spaces. Consider some x ∈πn+k(Sn). Realize x by a smooth map f ∶ Sn+k → Sn such that f(∞) = ∞ and such thatf is transverse to p (such maps clearly exist). Then f−1(p) is a smooth k-manifold inRn+k ⊂ Sn+k. Moreover, our fixed isomorphism (TRn)p ≅ Rn lifts to a framing on f−1(p).We define Ψ(x) to be the element of Ωfr

k (Rn+k) corresponding to f−1(p). Once we have

shown in the next two steps that this does not depend on the various choices we have made,it will be easy to see that Φ is a homomorphism.

Step 2. We prove that Φ does not depend on the choice of f .

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Let f ′ ∶ Sn+k → Sn be another smooth map such that f ′(∞) = ∞ and such that f ′ istransverse to p. We can then choose a homotopy F ∶ Sn+k × [0, 1]→ Sn from f to f ′ whichis transverse to p and satisfies F (∞, t) =∞ for t ∈ [0, 1]. The preimage F−1(p) is a framedcobordism from f−1(p) to (f ′)−1(p).

Step 3. We prove that Φ does not depend on the choice of p or the orientation-preservingisomorphism (TRn)p ≅ Rn.

Consider another q ∈ Rn together with an orientation-preserving isomorphism (TRn)q ≅Rn. We allow the possibility that q = p. There exists a compactly supported diffeomorphismτ ∶ Rn → Rn with the following properties.

• We have τ(p) = q.

• The compositionRn ≅ (TRn)p

τ∗Ð→ (TRn)q ≅ Rn

is the identity.

• The map τ is isotopic to the identity through an isotopy whose support is compact.Such a diffeomorphism is easy to construct. Letting τ ∶ Sn → Sn be the extension to the one-point compactification, the map τ is smooth and isotopic to the identity through an isotopythat fixes ∞. It follows that τ f ∶ Sn+k → Sn realizes x ∈ πn+k(Sn). The key observation isthat τ f is transverse to q and satisfies (τ f)−1(q) = f−1(p), where the indicated preimagesare given the framings coming from the given isomorphisms (TRn)q ≅ Rn and (TRn)p ≅ Rn.The desired independence follows.

Convention. Since we have shown that p can be arbitrary, from this point onwards wewill take p = 0 ∈ Rn+k ⊂ Sn+k.

Step 4. We prove that Φ is surjective.

Consider a framed k-manifold Mk in Rn+k. Let U ⊂ Rn+k be a tubular neighborhood ofMk. The framing induces a diffeomorphism U ≅Mk×Rn; let g ∶ U → Rn be the compositionof this diffeomorphism with the projection Mk ×Rn → Rn. Regarding Sn = Rn ∪ ∞ andSn+k = Rn+k ∪ ∞, we can then define a map f ∶ Sn+k → Sn via the formula

f(x) =⎧⎪⎪⎨⎪⎪⎩

g(x) if x ∈ U,

∞ otherwise.

It is then clear that Φ takes the element of πn+k(Sn) associated to f to Mk. The aboveconstruction of f is called the Pontryagin–Thom construction.

Step 5. We prove that Φ is injective.

Consider x ∈ πn+k(Sn) such that Φ(x) = 0. As in the definition of Φ, let f ∶ Sn+k → Sn

be a smooth map realizing x such that f(∞) = ∞ and such that f is transverse to p. SetMk = f−1(p). It is not hard to show that we can homotope f to the map obtained byapplying the Pontryagin–Thom construction to Mk. By assumption, there is a framedcobordism Ck+1 ⊂ Rn+k × [0, 1] from Mk to the empty manifold. Applying the Pontryagin–Thom construction to Ck+1 yields a homotopy from f to the constant map.

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Notation 3.2. If Mk is a framed k-manifold in Rn+k, then we will denote by ⟦Mk⟧ theassociated element of πn+k(Sn).

4 The Freudenthal suspension theoremAt this point in these notes, we will begin covering material that is hard to find in textbooks,so our proofs will become more complete. As a first illustration of the power of Theorem3.1, we will use it to prove the Freudenthal suspension theorem. This theorem will onlyuse the simplest type of framed cobordisms, namely those resulting from ambient isotopies,which we now define. Fix a framed k-manifold Mk in Rn+k. If ϕ ∶ Rn+k → Rn+k is adiffeomorphism, then we can define a framed k-manifold ϕ(Mk) in Rn+k in the obvious way.An ambient isotopy of Mk is a smooth function Φ ∶ Rn+k × [0, 1]→ Rn+k with the followingtwo properties.

• For all t ∈ [0, 1], the map Φ(⋅, t) is a diffeomorphism, which we will denote Φt.

• Φ0 = id.

The framed k-manifolds Mk and Φ1(Mk) are then framed cobordant via the cobordism

Ck+1 = (Φt(x), t) ∣ x ∈Mk, t ∈ [0, 1] ⊂ Rn+k × [0, 1],

which can be endowed with a framing in the obvious way. In a similar way, if Nk is obtainedfrom Mk by homotoping the framing, then Nk is framed cobordant to Mk.

We now define the suspension homomorphism E ∶ πn+k(Sn) → πn+k+1(Sn+1). Considera framed k-manifold Mk in Rn+k with framing f ∶Mk ×Rn → NRn+k/Mk . Define E(Mk) tobe the following framed k-manifold in Rn+k+1.

• The manifold underlying E(Mk) is the image of Mk under the standard inclusionRn+k Rn+k+1.

• The framing E(f) ∶ E(Mk) ×Rn+1 → NRn+k+1/E(Mk) is as follows. Let

a1, . . . , an+1 ⊂ Rn+1 and b1, . . . , bn+k+1 ⊂ Rn+k+1

be the standard bases. Then E(f) is defined via the formulas

E(f)(x, ai) = f(x, ai) (x ∈ E(Mk), 1 ≤ i ≤ n)

andE(f)(x, an+1) = bn+k+1 (x ∈ E(Mk)),

where we are identifying E(Mk) with Mk, the vector space Rn with its image in Rn+1,and the vector space Rn+k with its image in Rn+k+1.

This respects the relation of framed cobordism, and thus induces a map E ∶ Ωfrk (R

n+k) →Ωfr

k (Rn+k+1). By Theorem 3.1, this is the same as a map E ∶ πn+k(Sn)→ πn+k+1(Sn+1).

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Remark. Recall that we have a based suspension functor Σ that satisfies ΣSp ≅ Sp+1. Itis an easy exercise to see that the above map E ∶ πn+k(Sn) → πn+k+1(Sn+1) is the same asthe map that takes a map Sn+k → Sn to the suspended map

Sn+k+1 ≅ ΣSn+k Ð→ ΣSn ≅ Sn+1.

The letter E stands for the German word for suspension, namely Einhängung.

The following theorem is due to Freudenthal [3].

Theorem 4.1 (Freudenthal suspension theorem). The suspension homomorphism

E ∶ πn+k(Sn)→ πn+k+1(Sn+1)

is surjective for n ≥ k + 1 and an isomorphism for n ≥ k + 2.

Proof. We will prove that E is surjective for n ≥ k+1; to prove that it is an isomorphism forn ≥ k+2, simply run the proof below for a framed cobordism between two framed k-manifoldin Rn+k+1 (this requires n ≥ k + 2 since the framed cobordism is (k + 1)-dimensional). Anelement of πn+k+1(Sn+1) can be written in the form ⟦Mk⟧ for some framed k-manifold inRn+k+1. We will prove that Mk can be isotoped such that it is of the form E(Nk) for someframed k-manifold in Rn+k. This will be accomplished in two steps.

Step 1. We isotope Mk such that it lies in Rn+k ⊂ Rn+k+1.

The usual proof of the Whitney embedding theorem (see, e.g., [6, Theorem 1.3.5]) showsthat for a generic nonzero vector ζ in Rn+k+1, the projection of Mk onto the orthogonalcomplement of ζ is a smooth embedding. In particular, we can find such a ζ = (ζ1, . . . , ζn+k+1)with ζn+k+1 > 0. Let

X = x ∈ Rn+k+1 ∣ there exists s ∈ R such that x + sζ ∈Mk,

and define ϕ ∶X → R by letting ϕ(x) for x ∈X be the (n+ k + 1)st coordinate of the uniquepoint of Mk of the form x + sζ for some s ∈ R. The set X is closed and ϕ is smooth, so wecan extend ϕ to a smooth function ϕ ∶ Rn+k+1 → R. For 0 ≤ t ≤ 1, define Ψt ∶ Rn+k+1 → Rn+k+1

via the formulaΨt(x) = x − tϕ(x)

ζn+k+1ζ (x ∈ Rn+k+1).

Observe that each Ψt is a diffeomorphism of Rn+k+1, that Ψ0 = id, and that Ψ1 takes Mk

to a submanifold of Rn+k, as desired.

Step 2. We isotope the framing on Mk such that Mk = E(Nk) for some framed k-manifoldin Rn+k.

Let f ∶Mk ×Rn+1 → NRn+k+1/Mk be the framing. We will regard each fiber of NRn+k+1/Mk

as a subspace of Rn+k+1. Letting

a1, . . . , an+1 ⊂ Rn+1 and b1, . . . , bn+k+1 ⊂ Rn+k+1

be the standard bases, our goal is to construct a family ft ∶ Mk × Rn+1 → NRn+k+1/Mk offramings of Mk with the following properties.

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• f0 = f.

• f1(x, an+1) = bn+k+1 for all x ∈Mk.

• f1(x, ai) ⊂ Rn+k ⊂ Rn+k+1 for all x ∈Mk and 1 ≤ i ≤ n.

By Step 1, each fiber of NRn+k+1/Mk contains bn+k+1. Define ϕ ∶Mk → (Rn+1 ∖ 0) via theformula

f(x, ϕ(x)) = bn+k+1 (x ∈Mk).Since k ≤ n− 1 and (Rn+1 ∖ 0) is homotopy equivalent to Sn, the map ϕ is nullhomotopic.We can thus find a family ϕt ∶Mk → (Rn+1∖0) of maps such that ϕ1 = ϕ and ϕ0(x) = an+1for all x ∈Mk.

Next, for x ∈ Mk let µx be the inner product on Rn+1 obtained by pulling back thestandard inner product on NRn+k+1/Mk ⊂ Rn+k+1 via the map f(x, ⋅). The space of innerproducts on Rn+1 is contractible, so for x ∈Mk and 0 ≤ t ≤ 1 we can find an inner productµx,t on Rn+1 depending continuously on x and t such that µx,1 = µx and such that µx,0 isthe usual inner product on Rn+1.

Define B to be the vector bundle on Mk × [0, 1] whose fiber over (x, t) ∈Mk × [0, 1] isthe subspace of Rn+1 consisting of the µx,t-orthogonal complement of ϕt(x). For x ∈ Mk

and t = 0, we have ϕt(x) = an+1 and µx,t is the standard inner product on Rn+1, so therestriction of B to Mk ×0 is the trivial vector bundle (Mk ×0)×Rn. It follows that B itselfis a trivial vector bundle. We can therefore find linearly independent sections η1, . . . , ηn ofit such that ηi(x, 0) = ai for all x ∈Mk. The desired family of framing of Mk can then bedefined via the formula

ft(x,n+1∑i=1

ciai) = f(x, cn+1ϕt(x) +n

∑i=1

ciηi(x, t)) (x ∈Mk, c1, . . . , cn+1 ∈ R).

5 Framed circles and πn+1(Sn)In this section, we use geometric methods to show that

πn+1(Sn) =⎧⎪⎪⎨⎪⎪⎩

Z if n = 2,

Z/2 if n ≥ 3.

This is proven in Theorem 5.5 below. We begin with the following theorem of Hurewicz–Steenrod [10].

Theorem 5.1. For k ≥ 3, we have πk(S2) ≅ πk(S3).

Proof. Recall that the Hopf fibration is the fiber bundle

S1 S3 → S2

arising from the restriction of the map (C2)∗ → CP1 ≅ S2 to S3 ⊂ (C2)∗. The associatedlong exact sequence of homotopy groups contains the segment

πk(S1)→ πk(S3)→ πk(S2)→ πk−1(S1).

Since πk(S1) = πk−1(S1) = 0 for k ≥ 3, we deduce that πk(S2) ≅ πk(S3).

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Since π3(S3) ≅ Z, this implies the following theorem of Hopf [8].

Corollary 5.2. We have π3(S2) = Z.

To deal with πn+1(Sn) for n ≥ 3, we will construct an invariant of framed circles in Rn+1.This will require an understanding of the fundamental group of SLn(R).

Lemma 5.3. We have

π1(SLn(R)) =⎧⎪⎪⎨⎪⎪⎩

Z if n = 2,

Z/2 if n ≥ 3.

Proof. The Lie group SLn(R) deformation retracts onto its maximal compact subgroupSOn(R) (for instance, via the Gram–Schmidt process), so it is enough to prove that

π1(SOn(R)) =⎧⎪⎪⎨⎪⎪⎩

Z if n = 2,

Z/2 if n ≥ 3.

The cases n = 2 and n = 3 follow from the fact that SO2(R) ≅ S1 and SO3(R) ≅ RP3. Forthe cases n ≥ 4, the long exact sequence associated to the fiber bundle

SOn−1(R) SOn(R)→ Sn−1

contains the segment

π2(Sn−1)→ π1(SOn−1(R))→ π1(SOn(R))→ π1(Sn−1).

Since n ≥ 4, we have π2(Sn−1) = π1(Sn−1) = 0, so π1(SOn(R)) ≅ π1(SOn−1(R)) and thelemma follows.

Now consider a framed 1-manifold M1 in Rn+1 with n ≥ 2. Let f ∶M1 ×Rn → NRn+1/M1

be the framing. Define a function

ρM1 ∶M1 → SLn+1(R)

as follows. Recall that each fiber of NRn+1/M1 is a subspace of Rn+1. Let

a1, . . . , an ⊂ Rn

be the standard basis. For x ∈M1, there exists a unique vector τ(x) ∈ Rn+1 in the tangentspace to M1 at x such that the (n + 1) × (n + 1) matrix

ρM1(x) ∶= (τ (x) f (x, a1) f (x, a2) ⋯ f (x, an))

has determinant 1.Orienting M1 in the direction of τ(x), we can now define

P(M1) = (ρM1)∗([M1]) ∈ H1(SLn+1(R);Z) ≅ Z/2.

This is almost an invariant of the framed cobordism class of M1. The only problem is that ifM1 is the trivially framed unit circle in R2 ⊂ Rn+1, then P(M1) = 1 (easy exercise!); however,

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Figure 1: On the left is the transition from M1t0−ϵ to M1

t0+ϵ. On the right is the oriented loop L.

since ⟦M1⟧ = 0, we want the answer to be 0. This motivates the following definition: if M1

is a framed 1-manifold in Rn+1 with n ≥ 2, then

P(M1) = P(M1) + p ∈ Z/2,

where p is the number of components of M1. We then have the following lemma, whichimplies that we can define P(x) for x ∈ πn+1(Sn) by setting P(x) = P(M1), where M1 is aframed 1-manifold in Rn+1 with ⟦M1⟧ = x.

Lemma 5.4. Fix some n ≥ 2. Let M10 and M1

1 be framed 1-manifolds in Rn+1 which areframed cobordant. Then P(M1

0 ) = P(M11 ).

Remark. If n = 1, then we could define an integer valued invariant P(M1) of framed 1-manifolds in Rn+1 using the above procedure. However, this would be of little use since thisinvariant would in fact always vanish (easy exercise!). This reflects the fact that πk(S1) = 0for k ≥ 2.

Proof of Lemma 5.4. Let C2 ⊂ Rn+1 × [0, 1] be a framed cobordism between M10 × 0 and

M11 × 1. Perturbing C2 slightly, we can assume that the projection Rn+1 × [0, 1] → [0, 1]

restricts to a Morse function π ∶ C2 → [0, 1] whose critical points project to distinct criticalvalues. For all regular values t ∈ [0, 1] of π, define M1

t = π−1(t) and C2≤t = π−1([0, t]).

Regarding M1t as a 1-dimensional submanifold of Rn+1, we can endow M1

t with a framingon its normal bundle by orthogonally projecting the framing on C2 to Rn+1; with thisdefinition, M1

0 and M11 agree with the framed manifolds we began with. As t increases

from 0 to 1, it is clear that P(M1t ) can only change when t passes through a critical value.

Letting t0 ∈ [0, 1] be a critical value and ϵ > 0 be small enough such that the only criticalvalue in [t0 − ϵ, t0 + ϵ] is t0, we must prove that P(M1

t0−ϵ) = P(M1t0+ϵ). Let x ∈ C2 be the

critical point with π(x) = t0. There are three cases.

• The 2-manifold C2t0+ϵ is obtained from C2

t0−ϵ by attaching a 0-handle. In other words,the point x is a local minimum for π. In this case, M1

t0+ϵ differs from M1t0−ϵ in that

a component V appears. It is clear that P(V ) = 1, so P(V ) = 0 and P(M1t0−ϵ) =

P(M1t0+ϵ).

• The 2-manifold C2t0+ϵ is obtained from C2

t0−ϵ by attaching a 1-handle. In other words,the point x is a saddle point for π. The change from M1

t0−ϵ to M1t0+ϵ is as depicted

in Figure 1. As is clear from that picture, M1t0+ϵ has either one more or one fewer

component than M1t0−ϵ (this depends on whether the two points that are coming

together lie on the same component or on different components). Also, the homology

10

classes P(M1t0−ϵ) ∈ H1(SLn+1(R)) ≅ Z/2 and P(M1

t0+ϵ) ∈ H1(SLn+1(R)) ≅ Z/2 differ byP(L), where L is the loop shown in Figure 1. The loop L bounds a disc on the framedsurface C2, so P(L) = 1. In summary, P(M1

t0+ϵ) ∈ Z/2 is obtained from P(M1t0−ϵ) ∈ Z/2

by adding 1 (the change in the number of components) and then adding 1 again (thechange coming from L). The net effect is that P(M1

t0−ϵ) = P(M1t0+ϵ), as desired.

• The 2-manifold C2t0+ϵ is obtained from C2

t0−ϵ by attaching a 2-handle. In other words,the point x is a local maximum for π. This case is similar to that of a 0-handle.

We now define a special case of the J-homomorphism, which was originally defined byWhitehead [16] following work of Hopf [8]. The homomorphism we need is of the form

Jn ∶ π1(SLn(R))→ πn+1(Sn)

and can be defined as follows. Let L2 ⊂ R2 be the unit circle, oriented in the clockwisedirection. We have NR2/L2

≅ L2 ×R. Endowing NR2/L2with the inner product coming from

the standard inner product on R2, we frame L2 with the framing whose value on x ∈ L2 is theoutward facing unit vector. For n ≥ 3, let Ln ⊂ Rn be the result of stabilizing the framed 1-manifold L and let f ∶ Ln×Rn−1 → NRn/Ln

be the framing. For a map θ ∶ Ln+1 → SLn(R), theframing on Ln+1 obtained by twisting f by θ is the framing whose value on (x, v) ∈ Ln+1×Rn isf(x, θ(x)(v)). Up to homotopy, this only depends on the element of π1(SLn(R)) determinedby θ, so we will speak of twisting f by elements of π1(SLn(R)). We will denote by Ln+1(θ)the framed 1-manifold in Rn+1 obtained by twisting the framing of Ln+1 by an elementθ ∈ π1(SLn(R)). Then the J-homomorphism is defined via the formula

Jn(θ) = ⟦Ln+1(θ)⟧ (θ ∈ π1(SLn(R))).

It is easy to see that this is a homomorphism.

Remark. More generally, the J-homomorphism takes πk(SLn(R)) to πn+k(Sn).

We now prove the following theorem, which is due to Freudenthal [3].

Theorem 5.5. For all n ≥ 2, the J-homomorphism is an isomorphism. In particular,

πn+1(Sn) =⎧⎪⎪⎨⎪⎪⎩

Z if n = 2,

Z/2 if n ≥ 3.

Proof. That the J-homomorphism is an isomorphism for n = 2 follows easily from the proofof Corollary 5.2. Assume now that n ≥ 3. Let E ∶ π3(S2)→ πn+1(Sn) be the map obtained byiterating the suspension map. The Freudenthal suspension theorem (Theorem 4.1) impliesthat E is surjective. Next, let F ∶ π1(SL2(R)) → π1(SLn(R)) be the map induced by theusual inclusion SL2(R) SLn(R). The proof of Lemma 5.3 shows that F is surjective.Combining the evident commutative diagram

π1(SL2(R))J2ÐÐÐ→ π3(S2)

×××ÖF

×××ÖE

π1(SLn(R))JnÐÐÐ→ πn+1(Sn)

11

with the above facts, we see that Jn ∶ π1(SLn(R)) → πn+1(Sn) is surjective. Letting θ ∈π1(SLn(R)) ≅ Z/2 be a generator, it is clear from the construction that P(Jn(θ)) = 1, soJn ∶ π1(SLn(R))→ πn+1(Sn) is also injective, as desired.

Remark. If M1 is a framed 1-manifold in R3, then P(M1) ∈ Z/2 is the reduction modulo2 of ⟦M1⟧ ∈ π3(S2) ≅ Z.

6 Linking numbersBefore we can calculate πn+2(Sn), we need four sections of preliminaries. This sectionreviews the theory of linking numbers.

Let Ak and Bl be disjoint oriented closed submanifolds of Rk+l+1 whose dimensions arek and l, respectively. We can then define a smooth function f ∶ Ak × Bl → Sk+l via theformula

f(a, b) = a − b

∥a − b∥∈ Sk+l ⊂ Rk+l+1.

The linking number of Ak and Bl, denoted Link(Ak, Bl), is the degree of f .The following lemma shows the effect on their linking number of flipping Ak and Bl.

Lemma 6.1. Let Ak and Bl be disjoint oriented closed submanifolds of Rk+l+1 whose di-mensions are k and l, respectively. Then

Link(Ak, Bl) = (−1)(k+1)(l+1)Link(Bl, Ak).

Proof. Let f ∶ Ak × Bl → Sk+l be as in the first definition of the linking pairing, soLink(Ak, Bl) is the degree of f . Let λ ∶ Ak × Bl → Bl × Ak flip the two factors and letµ ∶ Sk+l → Sk+l be the antipodal map. Then λ has degree (−1)kl and µ has degree (−1)k+l+1.The number Link(Bl, Ak) is the degree of µ f λ, so

Link(Bl, Ak) = (−1)kl(−1)k+l+1Link(Ak, Bl)= (−1)k+l+kl+1Link(Ak, Bl)

= (−1)(k+1)(l+1)Link(Ak, Bl).

Next, we want to show that linking numbers are unchanged by certain kinds of cobor-disms. As a preliminary to this, we need the following result.

Lemma 6.2. Let Xn+1 be a compact oriented (n + 1)-dimensional manifold with boundaryand let g ∶Xn+1 → Sn be a smooth map. Then the degree of g∣∂Xn+1 is 0.

Proof. Let p ∈ Sn be a regular value of g. The inverse image g−1(p) is then a properlyembedded 1-dimensional submanifold of Xn+1. Each component of g−1(p) is either a circleor an arc connecting two points of (g∣∂Xn+1)−1(p) whose signs are opposite. Because of this,all points of (g∣∂Xn+1)−1(p) whose sign is positive can be paired with points whose sign isnegative, so the degree of g∣∂Xn+1 is 0.

The following lemma is the promised result about cobordisms.

12

Lemma 6.3. Let Ak+1 and Bl+1 be disjoint properly embedded oriented compact submani-folds of Rk+l+1×[0, 1] whose dimensions are k and l, respectively. For i = 0, 1, let Ak+1

i (resp.Bl+1

i ) be the intersection of Ak+1 (resp. Bl+1) with Rk+l+1 × i. Give Ak+10 the orientation

coming from Ak+1 and Ak+11 the opposite orientation, and similarly for Bl+1

0 and Bl+11 . Then

Link(Ak+10 , Bl+1

0 ) = Link(Ak+11 , Bl+1

1 ).

Proof. Deforming Ak+1 and Bl+1 slightly (while fixing their boundaries), we can assumethat that the projection π ∶ Ak+1 ⊔ Bl+1 → [0, 1] is a Morse function such that distinctcritical points map to distinct critical values. Breaking [0, 1] up into small segments, wesee that it is enough to prove the special case where the projection Ak+1 ⊔Bl+1 → [0, 1] hasa single critical point. Assume without loss of generality that this critical point lies in Bl+1.There is therefore an orientation-preserving diffeomorphism τ ∶ Ak+1

0 × [0, 1] → Ak+1 thattakes Ak+1

0 × 0 to Ak+10 . Letting π′ ∶ Rk+l+1 × [0, 1] → Rk+l+1 be the projection, we can then

define a smooth map F ∶ Ak+10 ×Bl+1 → Sk+l via the formula

F (a, b) = π′(τ(a, π(b))) − π′(b)∥π′(τ(a, π(b))) − π′(b)∥

∈ Sk+l ⊂ Rk+l+1.

The key properties of F are as follows.

• ∂(Ak+10 ×Bl+1) = (Ak+1

0 ×Bl+10 ) ⊔ (Ak+1

0 ×Bl+11 ).

• The restriction of F to Ak+10 ×Bl+1

0 has degree Link(Ak+10 , Bl+1

0 ).

• The restriction of F to Ak+10 ×Bl+1

1 has degree −Link(Ak+11 , Bl+1

1 ) (the negative signoccurs here for orientation reasons).

Applying Lemma 6.2 to to F , we conclude that

Link(Ak+10 , Bl+1

0 ) = Link(Ak+11 , Bl+1

1 ),

as desired.

7 The Hopf invariantWe now discuss the Hopf invariant H(x) ∈ Z of an element x ∈ π2n−1(Sn). This wasintroduced by Hopf in his paper [9]. Its main purpose for us is that (as we will show in§9) it is precisely the obstruction for improving the bounds in the Freudenthal suspensiontheorem. This will play an important role in our calculation of πn+2(Sn).

Consider an element x ∈ π2n−1(Sn). We will regard S2n−1 as R2n−1 ∪ ∞ and Sn asRn ∪ ∞. Fix two distinct points p, q ∈ Rn. Represent x as a smooth map ϕ ∶ S2n−1 → Sn

such that ϕ(∞) =∞ and such that both p and q are regular values of ϕ (such maps clearlyexist). Observe that

ϕ−1(p), ϕ−1(q) ⊂ R2n−1 ⊂ S2n−1

are disjoint (n − 1)-dimensional oriented submanifolds of R2n−1. The Hopf invariant of x,denoted H(x), is Link(ϕ−1(p), ϕ−1(q)).

13

Lemma 7.1. The Hopf invariant does not depend on the choice of ϕ or on the choice of pand q.

Proof. We begin by proving that H(x) does not depend on the choice of ϕ. Let ϕ′ ∶ S2n−1 →Sn be another smooth map representing x such that ϕ′(∞) =∞ and such that both p andq are regular values of ϕ′. There thus exists a smooth map F ∶ S2n−1 × [0, 1]→ Sn such thatF (⋅, 0) = ϕ and F (⋅, 1) = ϕ′. Deforming F slightly (but fixing F (⋅, 0) and F (⋅, 1)), we canassume that both p and q are regular values of F , and hence that F−1(p) and F−1(q) aresubmanifolds of R2n−1 × [0, 1]. Applying Lemma 6.3 to these submanifolds, we deduce that

Link(ϕ−1(p), ϕ−1(q)) = Link((ϕ′)−1(p), (ϕ′)−1(q)),

as desired.We now check that H(x) does not depend on the choice of p and q. Let p′, q′ ∈ Rn be

two distinct points. There exists a compactly supported diffeomorphism τ ∶ Rn → Rn withthe following properties.

• We have τ(p) = p′ and τ(q) = q′.

• The map τ is isotopic to the identity through an isotopy whose support is compact.

Such a diffeomorphism is easy to construct. Letting τ ∶ Sn → Sn be the extension to theone-point compactification, the map τ is smooth and isotopic to the identity through anisotopy that fixes ∞. It follows that τ ϕ ∶ S2n−1 → Sn realizes x ∈ π2n−1(Sn). The keyobservation is that τ ϕ is transverse to p′ and q′, and satisfies

(τ ϕ)−1(p′) = ϕ−1(p) and (τ ϕ)−1(q′) = ϕ−1(q).

The desired independence follows.

As we said above, we will later prove that the Hopf invariant is the obstruction toimproving the range in the Freudenthal suspension theorem. The following is a cheapcondition that ensures that the Hopf invariant vanishes.

Lemma 7.2. If n is even, then H(x) = 0 for all x ∈ π2n−1(Sn).

Proof. Immediate from Lemma 6.1.

The final result in this section shows how to calculate H(x) from a framed manifoldrepresenting x.

Lemma 7.3. Let Mn−1 be a framed (n − 1)-dimensional manifold in R2n−1 with framingf ∶Mn−1×Rn → NR2n−1/Mn−1. Regarding the fibers of NR2n−1/Mn−1 as n-dimensional subspacesof R2n−1, for v ∈ Rn define

Mn−1(v) = x + f(x, v) ∣ x ∈Mn−1.

Then for v ∈ Rn sufficiently close to 0, the set Mn−1(v) is an (n−1)-dimensional submanifoldof R2n−1 which is disjoint from Mn−1 and

H(⟦Mn−1⟧) = Link(Mn−1, Mn−1(v)).

14

Figure 2: On the left is a good 1-handle H for a framed 1-manifold M1 together with M1(H). Onthe right is a good 1-handle H for a framed 2-manifold M2 together with M2(H).

Proof. It is obvious that for v ∈ Rn sufficiently close to 0, the set Mn−1(v) is an (n − 1)-dimensional submanifold of R2n−1 which is disjoint from Mn−1. The indicated formulafollows from the fact that the sets Mn−1 and Mn−1(v) are pullbacks of regular values underthe map given by the Pontryagin–Thom construction.

8 Framed surgeryOur promised improvement to the Freudenthal suspension theorem will require more dras-tic framed cobordisms than the ambient isotopies used to prove the ordinary Freudenthalsuspension theorem. The basic tool for constructing these framed cobordisms is framedsurgery. Thankfully we will only need the simplest case of this.

Let Mk be a framed k-manifold in Rn+k. The framing provides an orientation on thenormal bundle NRn+k/Mk . The direct sum TMk ⊕NRn+k/Mk is the pullback to Mk of TRn+k ,which is also oriented, so TMk acquires an orientation. We will call this orientation on Mk

the orientation induced by the framing.For all p ≥ 0, let Dp denote a p-dimensional closed disc. Observe that ∂(Dp ×Dq) =

(Sp−1×Dq)∪(Dp×Sq−1). The subspaces Sp−1×Dq and Dp×Sq−1 of ∂(Dp×Dq) meet alongSp−1 × Sq−1.

A good 1-handle for Mk is a (k + 1)-dimensional oriented submanifold H of Rn+k withthe following properties (see Figure 2 for examples).

• H ≅D1 ×Dk.

• H ∩Mk = S0 ×Dk.

• The orientation of S0 ×Dk ⊂ ∂(D1 ×Dk) is the same as the orientation this manifoldacquires as a codimension 0 submanifold of Mk (which is given the orientation inducedby the framing).

Let Y = S0 ×Dk ⊂ ∂H and Z =D1 ×Sk−1 ⊂ ∂H. As in Figure 2, let Mk(H) be the orientedk-dimensional submanifold of Rn+k obtained by deleting Int(Y ) ⊂H ∩Mk from Mk, addingZ, and then “smoothing the corners”.

Lemma 8.1. Let Mk be a framed k-manifold in Rn+k and let H ⊂ Rn+k be a good 1-handlefor Mk. Then there exists a framing on Mk(H) such that Mk is framed cobordant toMk(H).

15

Figure 3: On the left is a 1-manifold with a good 1-handle (in bold). On right right is the cobordismconstructed in Lemma 8.1; the handle has been “pushed” down.

Proof. Let t ∈ [0, 1] denote the final coordinate in Rn+k × [0, 1]. As is shown in Figure 3,we can obtain a cobordism Ck+1 ⊂ Rn+k × [0, 1] between Mk and Mk(H) via the followingprocedure (which results in a manifold with corners that should be smoothed).

• Begin withCk+1 ∶= (Mk ∖ Int (Y )) × [0, 1].

• Glue H to Ck+1 in the following way, which one can think of as “pushing” H ×1 downinto Rn+k × [0, 1].

– The submanifold Y of ∂H = Y ∪Z is at height t = 1.– Let U ⊂ Z be an open collar neighborhood of ∂Z. Then the submanifold Z ∖ U

of ∂H is at height t = 0.– The rest of H lies in Rn+k × (0, 1).– The submanifold H of Ck+1 is orthogonal to the planes t = 0 and t = 1.

All that remains is to construct a framing on Ck+1. We already have a framing on Ck+1

and on Y ⊂ ∂H which we must extend over H. To deal with this extension, we will useobstruction theory (see [2, Chapter 7] and [5, p. 415] for two different expositions of this;the point of view of [2, Chapter 7] is more elementary). The first obstruction to extendingour framing over H lies in

H1(H, Y ; π0(GLn(R))) ≅ H1(H, Y ;Z/2) = Z/2.

This is generated by an arc connecting the two components of Y . Our assumption concerningthe orientations in the definition of “good 1-handle” implies that this first obstructionvanishes. Since H is contractible and Y ⊂ ∂H consists of two contractible components,there are no higher obstructions and the desired extension exists.

9 Improving the Freudenthal suspension theoremRecall that Theorem 4.1 (the Freudenthal suspension theorem) says that the suspensionmap πn+k(Sn) → πn+k+1(Sn+1) is surjective for n ≥ k + 1 and an isomorphism for n ≥ k + 2.These bounds are sharp in general, i.e. the suspension map π2n(Sn) → π2n+1(Sn+1) neednot be surjective. The following theorem (proved by Freudenthal in [3], the same paperthat contains the usual Freudenthal suspension theorem) says that the image is exactly theset of elements whose Hopf invariant is 0.

16

Theorem 9.1. The image of the suspension homomorphism E ∶ π2n(Sn)→ π2n+1(Sn+1) isx ∈ π2n+1(Sn+1) ∣ H(x) = 0.

Proof. We first prove that H(E(y)) = 0 for all y ∈ π2n(Sn). Write y = ⟦P n⟧, where P n is aframed n-dimensional manifold in R2n. Let b1, . . . , b2n+1 be the standard basis for R2n+1.Lemma 7.3 implies that H(E(y)) = Link(E(P n), E(P n) + b2n+1). This linking number isclearly 0.

Now consider x ∈ π2n+1(Sn+1) such that H(x) = 0. We will prove that x is in the imageof E. Write x = ⟦Mn⟧, where Mn is a framed n-dimensional manifold in R2n+1. Our goalis to prove that Mn is framed cobordant to E(Nn), where Nn is a framed n-dimensionalmanifold in R2n. Just like the proof of Theorem 4.1 (the Freudenthal suspension theorem),this will be accomplished in two steps.

Step 1. We find a framed cobordism from Mn to a submanifold of R2n ⊂ R2n+1.

We remark that this step does not use our assumption about the Hopf invariant. Itfollows from the proof of the Whitney embedding theorem (see [6, Theorem 1.3.5] and thediscussion on [6, p. 27]) that for a generic nonzero vector ζ in R2n+1, the projection ofMn onto the orthogonal complement of ζ is an immersion (not necessarily an embedding!)without triple points. Let ζ = (ζ1, . . . , ζ2n+1) be such a vector with ζ2n+1 > 0. We will provethat Mn can be modified by a framed cobordism such that the projection of Mn onto theorthogonal complement of ζ is an embedding. We will then be able to follow Step 1 of theproof of Theorem 4.1 line-for-line and deduce the desired result.

The fact that the projection is an immersion implies that there are finitely many linesparallel to ζ that pass through more than one point of Mn. Let these lines be ℓ1, . . . , ℓm.The fact the projection has no triple points implies that each ℓi passes through exactly twopoints of Mn. If m = 0, then we are done, so assume that m ≥ 1. Let ℓ1 ∩Mn = p, q.We can then find a neighborhood U of ℓ1 that does not intersect any other ℓi and anorientation-preserving diffeomorphism Ψ ∶ R2n+1 → U with the following properties.

• The image of tb2n+1 ∣ t ∈ R is ℓ1.

• The image of any line parallel to b2n+1 is a line parallel to ℓ1.

• SetV = v ∣ v ∈ ⟨b1, . . . , bn⟩ and W = v + b2n+1 ∣ v ∈ ⟨bn+1, . . . , b2n⟩.

Then Ψ(V ∪W ) =Mn ∩U .

• The orientation on V induced by the framing on Mn is defined by the ordered basisb1, . . . , bn.

• The orientation on W induced by the framing on Mn is opposite to that defined bythe ordered basis bn+1, . . . , b2n.

The image H under Ψ of the set

v + tb2n+1 ∣ t ∈ [0, 1], v =n

∑i=1

ci(cos(tπ

2)bi + sin(tπ

2)bn+i) for (c1, . . . , cn) ∈ Rn w/

n

∑i=1

c2i ≤ 1

17

is then a good 1-handle for Mn. Define Mn(H) as in §8, so by Lemma 8.1 we have thatMn is framed cobordant to Mn(H). Clearly ℓ1 is disjoint from Mn(H), the lines ℓ2, . . . , ℓm

each intersect Mn(H) exactly twice, and no other lines parallel to ζ intersect Mn(H) inmore than one place. We now replace Mn with Mn(H). Repeating this procedure, we cancomplete the proof.

Step 2. We isotope the framing on Mn such that Mn = E(Nn) for some framed n-manifoldin R2n.

This can be done exactly like Step 2 of the proof of Theorem 4.1. The only difference isthat we cannot use our dimension assumptions to prove that the map ϕ that appears thereis nullhomotopic. However, using the interpretation of the Hopf invariant in Lemma 7.3,our assumption that H(x) = 0 implies immediately that this map ϕ is nullhomotopic.

10 Framed surfaces and πn+2(Sn)We now turn to πn+2(Sn). Our goal is to prove Theorem 10.5 below, which asserts thatπn+2(Sn) = Z/2 for n ≥ 2. The key to this will be the construction of a Z/2-valued invariantof framed 2-manifolds in Rn+2.

Consider a framed 2-manifold M2 in Rn+2 with framing f ∶ M2 × Rn → NRn+2/M2 . Webegin by recalling some conventions introduced earlier. The framing endows NRn+2/M2 withan orientation. We will regard the fibers of the tangent bundle TM2 and the normal bundleNRn+2/M2 as subspaces of Rn+2, and we will endow them with the inner product obtained byrestricting the standard inner product on Rn+2. We will give M2 the orientation for whichthe direct sum decomposition TM2 ⊕NRn+2/M2 gives the standard orientation on Rn+2.

If L is an oriented 1-submanifold of M2, then we can frame L as follows. For all x ∈ L,let n(x) be the unique unit tangent vector to M2 at x with the following properties.

• The vector n(x) is orthogonal to L.

• The vector n(x) points to the right of L (using the orientations on L and M2).

Also, let a1, . . . , an+1 be the standard basis for Rn+1. Our framing on L is then thefunction fL ∶ L ×Rn+1 → NRn+2/L defined via the formula

fL(x, v + can+1) = f(x, v) + cn(x) (v ∈ Rn, c ∈ R).

We will call this the framing on L induced by M2.We now prove the following lemma.

Lemma 10.1. Consider a framed 2-manifold M2 in Rn+2. Let L and L′ be homologousoriented 1-submanifolds of M2. Endow L and L′ with the framings induced by M2. ThenL and L′ are framed cobordant as framed 1-manifolds in Rn+2.

Proof. Since L and L′ are homologous, there exists an oriented properly embedded surfaceΣ2 in M2 × [0, 1] such that

∂Σ2 = Σ2 ∩ (M2 × 0, 1) = (L × 0) ∪ (L′ × 1).

18

Figure 4: On the far left is our intersection pi. The second figure is a transitional one obtainedby just homotoping L′′. The third figure from the left is L′′′. The fourth figure is L(4); eliminatingthe intersection while keeping the framing from the ambient surface requires eliminating one “twist”from the framing.

For the reader who has not seen this kind of thing before, Σ can be obtained by taking animmersed surface with the above properties (which clearly exists) and then resolving theself-intersections (a collection of circles) to make it embedded. The 3-manifold M2 × [0, 1]in Rn+3× [0, 1] acquires a framing from M2, and using this we can put a framing on Σ2 justlike we did for L. This gives the desired framed cobordism.

By Lemma 10.1, we can define a function

qM2 ∶ H1(M2;Z)Ð→ Z/2

by setting qM2(x) = P(L), where L is an oriented 1-submanifold of M2 representing xendowed with the framing induced by M2 and P(L) is the invariant of framed 1-manifoldsin Rn+2 introduced in §5. One’s first impulse might be that qM2 is a homomorphism; indeed,in [13] Pontryagin assumes that this holds and uses it to give a proof of the (false) theoremthat πn+2(Sn) = 0 for n ≥ 3. The correct statement is as follows. Let i(⋅, ⋅) be the Z/2-valued algebraic intersection pairing on H1(M2;Z/2). For x, y ∈ H1(M2;Z), we will leti(x, y) denote i(x, y), where x and y are the images in H1(M2;Z/2) of x and y, respectively.

Lemma 10.2. If M2 is a framed surface in Rn+2 with n ≥ 1, then

qM2(x + y) = qM2(x) + qM2(y) + i(x, y) (x, y ∈ H1(M2;Z)).

Proof. Let L and L′ be oriented 1-submanifolds of M2 realizing the homology classes x andy, respectively. Homotoping L and L′, we can assume that they intersect transversely. LetL ∩ L′ = p1, . . . , pk, so k ≡ i(x, y) modulo 2. Let L′′ be the result of taking L ∪ L′ andpushing one of the two strands meeting at each pi off the surface into Rn+2 as in Figure 4.a(it does not matter how you do this). Give L′′ the framing coming from the framings on Land L′ induced by M2. Thus

P(L′′) = P(L) +P(L′) = qM2(x) + qM2(y).

Let L′′′ be the result of modifying L′′ in a neighborhood of each pi as shown in Figure 4.b.As is clear from that Figure, L′′′ is framed cobordant to L′′ (indeed, it can be obtainedfrom L′′ by a sequence of surgeries on good 1-handles as in Lemma 8.1), so P(L′′′) = P(L′′).Finally, let L(4) be the result of “pulling tight” each self-intersection of L′′′ and giving theresult the framing coming from M2. From Figure 4, it is clear that this requires eliminatingone “twist” in the framing for each self-intersection, so

qM2(x + y) = P(L(4)) = P(L′′′) + k = qM2(x) + qM2(y) + i(x, y),

19

as desired.

Lemma 10.2 implies in particular that qM2(2x) = 2qM2(x) = 0 for all x ∈ H1(M2;Z).This implies that qM2 descends to a function

qM2 ∶ H1(M2;Z/2)Ð→ Z/2

satisfying

qM2(x + y) = qM2(x) + qM2(y) + i(x, y) (x, y ∈ H1(M2;Z/2)).

Such functions are known as quadratic refinements of the nondegenerate symmetric bilinearform i(⋅, ⋅) (this is symmetric since we are working over Z/2). Since we are working overZ/2, such quadratic refinements are classified by their Arf invariants, which we brieflydescribe (see, e.g., [15, Appendix] for details and proofs; a broader discussion of Arf’s workon this subject, of which the above is a tiny special case, is in [11]). A symplectic basis forH1(M2;Z/2) is a basis a1, b1, . . . , ag, bg such that

i(ai, bj) = δij and i(ai, aj) = i(bi, bj) = 0 (1 ≤ i, j ≤ g),

where δij is the Kronecker delta. Choosing such a symplectic basis, we then define the Arfinvariant of qM2 to be

Arf(qM2) =g

∑i=1

qM2(ai)qM2(bi) ∈ Z/2.

This does not depend on the choice of symplectic basis. The relevance of the Arf invariantfor us is the following result.

Lemma 10.3. Let M20 and M2

1 be framed 2-manifolds in Rn+2 which are framed cobordant.Then Arf(qM2

0) = Arf(qM2

1).

Proof. Let C3 ⊂ Rn+2×[0, 1] be a framed cobordism between M20 ×0 and M2

1 ×1. PerturbingC3 slightly, we can assume that the projection Rn+2 × [0, 1] → [0, 1] restricts to a Morsefunction π ∶ C3 → [0, 1] whose critical points project to distinct critical values. For allregular values t ∈ [0, 1] of π, define M2

t = π−1(t) and C3≤t = π−1([0, t]). Regarding M2

t as a2-dimensional submanifold of Rn+2, we can endow M2

t with a framing on its normal bundleby orthogonally projecting the framing on C3 to Rn+2; with this definition, M2

0 and M21

agree with the framed manifolds we began with. As t increases from 0 to 1, it is clear thatArf(qM2

t) can only change when t passes through a critical value. Letting t0 ∈ [0, 1] be a

critical value and ϵ > 0 be small enough such that the only critical value in [t0 − ϵ, t0 + ϵ]is t0, we must prove that Arf(qM2

t0−ϵ) = Arf(qM2

t0+ϵ). Let x ∈ C3 be the critical point with

π(x) = t0. There are four cases

• The 3-manifold C3≤t0+ϵ is obtained from C3

≤t0−ϵ by attaching a 0-handle. In other words,the point x is a local minimum for π. In this case, M2

t0+ϵ differs from M2t0−ϵ in that

a spherical component V appears. Since H1(V ;Z/2) = 0, this component contributesnothing to the Arf invariant and Arf(qM2

t0−ϵ) = Arf(qM2

t0+ϵ).

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Figure 5: On the left is the surface M2t0−ϵ, on the right is the surface M2

t0+ϵ. The “core” L of thenew tube is depicted.

• The 3-manifold C3≤t0+ϵ is obtained from C3

≤t0−ϵ by attaching a 1-handle. In this case,M2

t0+ϵ is obtained from M2t0−ϵ by attaching a tube as in Figure 5. Let L be the

oriented 1-submanifold of M2t0+ϵ forming the “core” of this tube, as shown in Figure

5. Endow L with the framing induced by M2t0+ϵ. The loop L bounds a disc in the

framed 3-manifold C3≤t0+ϵ. This disc can be framed (just like the cobordism in Lemma

10.1), so L is framed cobordant to the empty manifold and P(L) = 0. We now canchoose a symplectic basis a1, b1, . . . , ag, bg for H1(M2

t0+ϵ;Z/2) such that ag equalsthe homology class of L and a1, b1, . . . , ag−1, bg−1 comes from H1(M2

t0−ϵ;Z/2) in theobvious sense. We then have

Arf(qM2t0+ϵ) = (

g−1∑i=1

qM2t0−ϵ(ai) qM2

t0−ϵ(bi)) + qM2

t0+ϵ(ag)qM2

t0+ϵ(bg) = Arf(qM2

t0−ϵ) + 0,

as desired.

• The 3-manifold C3≤t0+ϵ is obtained from C3

≤t0−ϵ by attaching a 2-handle. This is similarto the case of a 1-handle (but in this case M2

t0−ϵ can be obtained from M2t0+ϵ by

attaching a tube rather than the other way around).

• The 3-manifold C3≤t0+ϵ is obtained from C3

≤t0−ϵ by attaching a 3-handle. In other words,the point x is a local maximum for π. This is similar to the case of a 0-handle.

Using the above invariant, we will prove the following lemma.

Lemma 10.4. For n ≥ 2, let νn ∶ Sn+1 → Sn represent the generator for πn+1(Sn) (seeTheorem 5.5). Then νn νn+1 ∶ Sn+2 → Sn represents a nontrivial element of πn+2(Sn) forall n ≥ 2.

Proof. Let us begin by recalling the construction of νn from Theorem 5.5. Let L2 ⊂ R2 bethe unit circle, oriented in the clockwise direction. We have NR2/L2

≅ L2 × R. EndowingNR2/L2

with the inner product coming from the standard inner product on R2, we frameL2 with the framing whose value on x ∈ L2 is the outward facing unit vector. For n ≥ 3, letLn ⊂ Rn be the result of stabilizing the framed 1-manifold L. Finally, let Ln ⊂ Rn be theresult of twisting the framing of Ln by a generator for π1(SLn−1(R)). Theorem 5.5 saysthat νn = ⟦Ln+1⟧.

As in the proof of Theorem 3.1, we can represent νn by the map fn ∶ Sn+1 → Sn

obtained by applying the Pontryagin–Thom construction to Ln+1. Regarding Sn as Rn∪∞,under this construction every point of Rn is a regular value for fn and Ln+1 ⊂ Rn+1 is theinverse image of 0 ∈ Rn (framed using the framing obtained by pulling back the standard

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frame at 0). From this, we see that 0 ∈ Rn ⊂ Sn is a regular value of fn fn+1. DefineTn = (fn fn+1)−1(0) ⊂ Rn+2 (framed again by pulling back the standard frame at 0), so Tn

is a framed 2-manifold in Rn+2. It follows from the proof of Theorem 3.1 that νnνn+1 = ⟦Tn⟧,so to prove the lemma it is enough to show that Arf(qTn) = 1.

It follows from its definition that Tn = f−1n+1(Ln+1). Moreover, for all x ∈ Ln+1 the inverse

image f−1n+1(x) is a circle. It follows that Tn is a circle bundle over a circle. Since Tn is

orientable, we deduce that Tn is a torus. We now make the following constructions.

• Define A = f−1n+1(p), where p is some point in Ln+1. Thus A is a 1-submanifold of Tn.

Orient A arbitrarily and endow A with the framing induced by Tn. Then since fn+1represents a generator for πn+2(Sn+1), it follows that P(A) = 1.

• Let V ⊂ Rn+2 be one of the fibers in the tubular neighborhood of Ln+2 used in thePontryagin–Thom construction. It follows that (fn+1)∣V is a diffeomorphism from Vto Rn+1 ⊂ Sn+1. Define B = ((fn+1)∣V )−1(Ln+1), so B is a 1-submanifold of Tn. OrientB arbitrarily and endow B with the framing induced by Tn. Then since fn representsa generator for πn+1(Sn), it follows that ⟦B⟧ ∈ πn+2(Sn+1) is a generator, and thusthat P(B) = 1.

Let a ∈ H1(Tn;Z/2) and b ∈ H1(Tn;Z/2) be the homology classes of A and B, respectively.Then a, b is a symplectic basis for H1(Tn;Z/2), so by definition we have

Arf(qTn) = qTn(a)qTn(b) = P(A)P(B) = 1,

as desired.

We now come to the main theorem of this section, which was proved independently byPontryagin [14] and Whitehead [17]

Theorem 10.5. For n ≥ 2, we have πn+2(Sn) = Z/2. It is generated by νn νn+1, where νm

is as in Lemma 10.4.

Proof. The proof is by induction on n. The base case n = 2 follows from the following facts.

• Theorem 5.1, which implies that π4(S2) ≅ π4(S3).

• Theorem 5.5, which says that π4(S3) ≅ Z/2.

• Lemma 10.4, which says that ν2 ν3 represents a nontrivial element of π4(S2).

Now assume that n ≥ 3. Let E ∶ π4(S2) → πn+2(Sn) be the composition of the suspensionmaps. Combining the ordinary Freudenthal suspension theorem (Theorem 4.1) with theimproved suspension theorem (Theorem 9.1) and Lemma 7.2, we see that E is surjective.Lemma 10.4 implies that πn+2(Sn) is not trivial, so E cannot be the zero map. We concludethat E must be an isomorphism and hence πn+2(Sn) = Z/2. Finally, Lemma 10.4 impliesthat νn νn+1 represents a generator of this group.

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Andrew PutmanDepartment of MathematicsRice University, MS 1366100 Main St.Houston, TX 77005andyp@math.rice.edu

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